The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=3 and x=0, and a root of multiplicity 1 at x=-3
Find a possible formula for P(x).
P(x)= 


Given the information, we can construct a polynomial \( P(x) \) of degree 5 based on the roots and their multiplicities.

1. **Root at \( x = 3 \) with multiplicity 2**: This gives the factor \( (x - 3)^2 \).
2. **Root at \( x = 0 \) with multiplicity 2**: This gives the factor \( x^2 \).
3. **Root at \( x = -3 \) with multiplicity 1**: This gives the factor \( (x + 3) \).

Since the polynomial has a leading coefficient of 1, we can multiply these factors directly.

### Constructing \( P(x) \):
\[
P(x) = (x - 3)^2 \cdot x^2 \cdot (x + 3)
\]

Expanding this expression is optional but often helpful. The polynomial is:

\[
P(x) = x^2 (x - 3)^2 (x + 3)
\]

This is a valid formula for \( P(x) \) that satisfies all given conditions.

Would you like me to expand this further, or do you have questions on the steps? 

Here are related questions that might deepen your understanding:

1. How do you find the degree of a polynomial given its roots and multiplicities?
2. How does the leading coefficient affect the end behavior of a polynomial?
3. What are other properties of a polynomial determined by its roots and multiplicities?
4. How would the formula change if one of the roots had a higher multiplicity?
5. How do we verify the correctness of this polynomial?

**Tip:** When constructing polynomials, always check that the sum of the multiplicities matches the desired degree.

The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=3 and x=0, and a root of multiplicity 1 at x=-3. Find a possible formula for P(x). P(x) = 

Discover how to construct a polynomial P(x) of degree 5 with roots at x = 3, x = 0, and x = -3, including specific multiplicities and leading coefficient 1. This solution covers polynomial functions, roots, and the role of multiplicities, ideal for students in Grades 10-12.

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